$ E = \left[\begin{array}{rrr}0 & 2 & 1 \\ -2 & 2 & 0\end{array}\right]$ $ F = \left[\begin{array}{rr}4 & 5 \\ -2 & 4 \\ 2 & -2\end{array}\right]$ What is $ E F$ ?
Because $ E$ has dimensions $(2\times3)$ and $ F$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ E F = \left[\begin{array}{rrr}{0} & {2} & {1} \\ {-2} & {2} & {0}\end{array}\right] \left[\begin{array}{rr}{4} & \color{#DF0030}{5} \\ {-2} & \color{#DF0030}{4} \\ {2} & \color{#DF0030}{-2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ E$ , with the corresponding elements in column $j$ of the second matrix, $ F$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ E$ with the first element in ${\text{column }1}$ of $ F$ , then multiply the second element in ${\text{row }1}$ of $ E$ with the second element in ${\text{column }1}$ of $ F$ , and so on. Add the products together. $ \left[\begin{array}{rr}{0}\cdot{4}+{2}\cdot{-2}+{1}\cdot{2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ E$ with the corresponding elements in ${\text{column }1}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{0}\cdot{4}+{2}\cdot{-2}+{1}\cdot{2} & ? \\ {-2}\cdot{4}+{2}\cdot{-2}+{0}\cdot{2} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ E$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{0}\cdot{4}+{2}\cdot{-2}+{1}\cdot{2} & {0}\cdot\color{#DF0030}{5}+{2}\cdot\color{#DF0030}{4}+{1}\cdot\color{#DF0030}{-2} \\ {-2}\cdot{4}+{2}\cdot{-2}+{0}\cdot{2} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{0}\cdot{4}+{2}\cdot{-2}+{1}\cdot{2} & {0}\cdot\color{#DF0030}{5}+{2}\cdot\color{#DF0030}{4}+{1}\cdot\color{#DF0030}{-2} \\ {-2}\cdot{4}+{2}\cdot{-2}+{0}\cdot{2} & {-2}\cdot\color{#DF0030}{5}+{2}\cdot\color{#DF0030}{4}+{0}\cdot\color{#DF0030}{-2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-2 & 6 \\ -12 & -2\end{array}\right] $